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Phy 303l - Kapoor(mk9499 – homework09 – Turner...

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Unformatted text preview: Kapoor (mk9499) – homework09 – Turner – (60230) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A proton is released from rest in a uniform electric field of magnitude 1 . 5 × 10 5 V / m di- rected along the positive x axis. The proton undergoes a displacement of 0 . 4 m in the di- rection of the electric field as shown in the figure. + + + + + + + + + + + + +------------- v A = 0 + . 4 m v 1 . 5 × 10 5 V / m A B Apply the principle of energy conserva- tion to find the amount of the kinetic energy gained after it has moved 0 . 4 m. Correct answer: 9 . 61306 × 10- 15 J. Explanation: The change in the potential energy of the proton is Δ U = q p Δ V = (1 . 60218 × 10- 19 C) (- 60000 V) =- 9 . 61306 × 10- 15 J . Conservation of energy in this case is Δ K + Δ U = K f- K i + Δ U = 0 . So K f- K i =- Δ U = 9 . 61306 × 10- 15 J . 002 10.0 points Two alpha particles (helium nuclei), each consisting of two protons and two neu- trons, have an electrical potential energy of 6 . 41 × 10- 19 J ....
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Phy 303l - Kapoor(mk9499 – homework09 – Turner...

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